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Q. A load of bricks with mass m1 = 14.0 kg hangs from one end of a rope that passes over a small, frictionless pulley. A counterweight of mass m2 = 27.8 kg is suspended from the other end of the rope. The system is released from rest. Use g = 9.80 m/s2 for the magnitude of the acceleration due to gravity.

a. What is the magnitude of the upward acceleration of the load of bricks?

b. What is the tension in the rope while the load is moving?

I started off by drawing a free body diagram for each of the two masses. Each mass had two forces acting on it in the y-direction (weight of the mass and the tension of the rope).

I also did two equations for the total forces acting on the two masses as follows:

Brick -> Fb-y = Tb - m1g

Counterweight -> Fc-y = Tc - m2g

I calculated the tension in the rope caused by m1 and m2. I assumed that the force that would pull the bricks (m1) upwards would be equal to the total tension in the rope and got the difference between the weight of m1 and m2. The difference ended up being 135.24 N.

I then used F = mass * acceleration to solve for the acceleration needed in part A and here is where I'm getting confused. When calculating the acceleration a = F/m, shouldn't the mass used in the equation be the mass of the object being accelerated by the force? Why is it then to get the correct answer I have to use the combined mass (14 + 27.8) kg and not just the mass of the object being moved (14 kg).

The correct answer that I calculated was 3.24 m/s2 while the incorrect answer using only the mass of the object that was moving upwards was 9.66 m/s2.

Part B:

Why is the tension in the rope not just the tension caused by m2 minus the tension caused by m1?

I, now, understand why to calculate the force needed to move the bricks upwards you must add the acceleration due to gravity and the acceleration of the object. Is it because in order for the mass to accelerate upwards with an acceleration of 3.24 m/s2 the object must overcome the gravitational acceleration?

And why is the combined mass not used in this second part of the question?

Thank you and sorry if I was not clear about what concepts confused me.

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  • $\begingroup$ Sorry I couldn't provide an example of the free body diagram as I am still new. $\endgroup$ – Abdul97 May 30 '15 at 22:59
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I think what you're missing here is that you're dealing with a model pulley (by model pulley, I mean that it's small and frictionless): this means that the tension in the rope on both sides is equal in magnitude. So in your notation, Tb = Tc. Let's call the tension T.

Then we just apply Newton's Second Law (the F = ma form) to both the bricks and counterweight. In both cases we have to take the direction in which the objects accelerate as positive.

The bricks move up, so that means the tension is positive and the weight of the bricks is negative. So for the bricks, we have: 14a = T - 14g.

The counterweight moves down, so that means the tension is negative and the weight of the counterweight is positive. So for the counterweight we have: 27.8a = 27.8g - T

If we simply add the equations together, we get: 41.8a = 13.8g

This gives a=3.24 metres per second squared, which is the right answer. That's part (a) done!

To do part (b) we just have to substitute the value for a (3.24) into either of the equations we derived using Newton's 2nd Law. Let's choose the first. We get: T = 14*3.24 + 14g = 180 newtons (to 2 significant figures).

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  • $\begingroup$ Thank you. So whenever a model pulley is used, tension in the rope is equal on both sides? But why did you add the two equations together? was it to just solve for acceleration? $\endgroup$ – Abdul97 May 30 '15 at 23:32
  • $\begingroup$ You're welcome. Yes, that's a property of model pulleys. And yes, we had two equations with unknowns a and T. We wanted to find a, so I eliminated T by adding the two equations together. $\endgroup$ – Mr. Bump May 30 '15 at 23:38

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